Question

An ideal solenoid, of radius R and n turns per unit length, has a current flowing through it. The current, I, varies with time, t, according to I = I0 + at where I0 and a are constants. A conducting ring of radius, r, is placed inside the solenoid with its axis coinciding with the axis of the solenoid. The ring has a resistance per unit length of H (in units of Ω/m).

(a) Use Lenz’s law to determine the direction of the current in the ring relative to the direction of the current in the solenoid.

(b) Calculate an expression for the induced current in the ring. (c) Repeat your calculation for the case when the radius of the ring is larger than the radius of the solenoid. (d) Make a plot of the current in the ring as a function of the radius of the ring. Include curves for r < R and r > R. Explain what the plot shows.

c) The ring is replaced by a conducting wire in the shape of a square. The side length of the square is l, and it is made of the same conducting material as the ring. The square is tilted by an angle, θ (the angle between the area vector for the solenoid and the area vector for the square is θ). Repeat your calculations in parts a) and b) to determine the current in the square. Include calculations for when the square is completely inside the solenoid and completely outside the solenoid only.

Answer 1

a) Lenz's law defines that induced emf tries to oppose its cause, so in order to do that a clockwise current will produce a field towards the left in the solinoid, and the current in the loop will generate a magnetic field which opposes this, so the loop will have a current going anti-clockwise, i.e. current will have a direction opposite to the solinoid's current.

b)

For inside the solinoid,

flux,

the rate of change of thiss flux,

this is the induced emf in the loop.

Resistance of the loop =

According to Ohm's Law, current in the loop,

__________________________

For a loop bigger than the solinoid,

(flux changes inside Solinoid only so the radius of the outer coil r is not used here, the solinoid's radius R is used)

According to Ohm's Law, current in the loop,

For inside , , so current follows linear growth.

For outside, , so current decreases hyperbolically.

c)

For the square, Area =

and, perimeter = 4l

so, For inside the solinoid,

flux,

so, rate of change of flux,

resistance of the loop = 4lH

so, current in the loop =

_______________________

for outside the solinoid,

so, current in the square loop outside =

(doesn't depend on )